Combustion Analysis- Empirical and molecular formula

In summary, the conversation was about finding the empirical and molecular formula for menthol, a substance found in mentholated cough drops. Through combustion, the masses of CO2 and H2O produced from a 0.1005 g sample of menthol were given. The correct approach was to find the mass of C, H, and O in the sample, and use their ratios to determine the empirical formula. The molecular formula was then found by using the molar mass of menthol. The resulting formula was C10H20O, with a molar mass of approximately 156 g/mol.
  • #1
Dustyneo
4
1

Homework Statement


Menthol, the substance we can smell
in mentholated cough drops, is composed of C, H, and O. A
0.1005-g sample of menthol is combusted, producing 0.2829 g
of CO2 and 0.1159 g of H2O. What is the empirical formula
for menthol? If menthol has a molar mass of 156 g/mol, what is its molecular formula?.
Book answer:
The empirical and molecular formulas are C 10H20O

2. Homework Equations

Whole-number multiple = molecular weight /empirical formula weight

The Attempt at a Solution



Cant arrive at the same answer. I started off using the masses of the combustion products, in the case of carbon

(0,2829 g CO2 ) (mol CO2 /44,0 g CO2) (mol C/mol CO2) =
0,00643 mol C
As for H and O, I got 0,129 mol and 0,00643 mol respectively. Then comparing the relative number of moles of each element

C: 0,00643/0,00643=1 H: 0,129/0,00643=20 O:0,00643/0,00643 = 1 lead me obtaining CH20O as the empirical formula. What am I doing wrong?
 
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  • #2
Please show us how you got the 0.129 mol for H.
 
  • #3
And 0.00643 mol for O.
 
  • #4
Made a calculation mistake for H, the corrected value is 0,0129

(0,1159 g H2O) (mol H2O/ 18,0 g H2O) (2 mol H/ mol H2O)= 0,0129 mol H

For O, I also noticed that the actual number is closer to 0,00644:
(0,1159 g H2O) (mol H2O/ 18,0 g H2O) ( mol O/ mol H2O) = 0,00644 mol O
 
  • #5
How do you know that all the O in the H2O, and none of that in the CO2, comes from the O in the compound rather than the oxygen in the air? You need to take the equation
aCxHyOz + bO2 → cCO2 + dH2O
and balance it.
 
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  • #6
To add to what mjc123 wrote, finding mass of oxygen in the compound is quite easy once you know mass of carbon and hydrogen (and actually doesn't require stoichiometry).
 
  • #7
mjc123 said:
How do you know that all the O in the H2O, and none of that in the CO2, comes from the O in the compound rather than the oxygen in the air? You need to take the equation
aCxHyOz + bO2 → cCO2 + dH2O
and balance it.

Ok. In order to fill the values for a, c and d, I did these calculations,
Finding the amount of menthol molecules,

(0,1005 g C10H20O ) (mol C10H20O /156,09 G C10H20O) (6,02 X 1023 C10H20O molecules / mol C10H20O ) = 3,88 x 1020 menthol molecules

CO2 molecules
(0,2829 g CO2) (mol CO2/44,0 g CO2 ) (6,02 X 1023 molecules CO2 /mol CO2) = 3,87 X 1021 CO2 molecules
H2O molecules
(0,1159 g H2O) (mol H2O /18,0 g)(6,02 x 1023 water molecules/mol H2O) = 3,88 x 1021 water molecules

Simplifying these number of molecules into the equation (number of water and CO2 molecules is 10 times the amount of menthol molecules) I get

C10H20O + bO2 → 10 CO2 + 10 H2O . Am I on the right track?
 
Last edited:
  • #8
That's cheating, because you are using the formula of menthol in your calculation.
Go back to what Borek said. You have 0.2829g CO2. What mass of C does this contain? It can only have come from the menthol, so this is the mass of C in 0.1005g menthol.
You have 0.1159g water. How much H does this contain? It can only have come from the menthol, so this is the mass of H in 0.1005g menthol.
Now you have the mass of C and H, what is the mass of O in 0.1005g menthol?
Having the mass fractions, convert them into mole fractions, and work out the simplest whole-number ratio.
 
  • #9
mjc123 said:
What mass of C does this contain? It can only have come from the menthol
How much H does this contain? It can only have come from the menthol

can only have come from the menthol
I think I grasped where the mistake was . So I proceeded to only get the mass amount for C and H.
(0,2829 g CO2)(mol CO2/44,0 g CO2)(mol C/mol CO2 )(12,0 g C/mol C) = 0,0772 g C
(0,1159 g H2O)(mol H2O/18,0 g H2O) (2 mol H/mol H2O)( 1 g H/mol H) = 0,0129 g H
Then substracting the addition of these masses from the menthol to get the amount of O:
(0,1005 g) - (0,0772 + 0,0129) = 0,0104 g O
Obtaining moles and comparing relative numbers
(0,0772 g C)(mol C/ 12,0 g C) = 0,00643 mol C
(0,0129 g H)(mol H/ g H)= 0,0129 mol H
(0,0104 g O)(mol O/16,0 g O)= 0,0006504 mol H

C: 0,00643/0,0006504≈ 10 H: 0,0129/0,0006504≈20 O: 0,0006504/0,0006504=1 giving C10H20O as the empirical formula, with a formula weight of aprox. 156 amu. Then using the relevant equation
156 amu/156 amu=1 indicates C10H20O also being the molecular formula. Thanks for the help.
 
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Related to Combustion Analysis- Empirical and molecular formula

1. What is combustion analysis?

Combustion analysis is a method used to determine the empirical and molecular formula of a compound by burning a sample of the compound in the presence of excess oxygen and measuring the amount of carbon dioxide and water produced.

2. How does combustion analysis work?

In combustion analysis, a sample of the compound is heated in a combustion chamber with a known amount of excess oxygen. The compound reacts with the oxygen to form carbon dioxide and water, which are then collected and measured. These measurements are used to calculate the empirical and molecular formula of the compound.

3. What is the difference between empirical and molecular formula?

The empirical formula is the simplest ratio of the elements present in a compound, while the molecular formula is the actual number of atoms of each element in a molecule of the compound.

4. Why is combustion analysis important?

Combustion analysis is important because it allows scientists to determine the chemical formula of a compound, which is essential for understanding its properties and behavior. It is also a crucial step in the process of synthesizing and identifying new compounds.

5. Are there any limitations to combustion analysis?

Yes, there are some limitations to combustion analysis. For example, it can only be used for compounds that contain carbon and hydrogen. It also assumes that all of the carbon in the compound is converted to carbon dioxide during the combustion process, which may not always be the case. Additionally, combustion analysis does not provide information about the arrangement of atoms in a molecule, only the ratio of elements.

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